[Math] Do the columns of the given matrix span $\mathbb{R}^3$

matricesspan

I was given the matrix:

$$A=\left(\begin{matrix}-4 & -7 & 1 & 2\\
0 & 0 & 3 & 8\\5 & -1 &1 & -4\end{matrix}\right)$$

I arranged them into an augmented matrix and added a 5th column of zeroes. I row reduced until I got

$$B=\left(\begin{matrix}1 & 0 & 0 & -2 & 0\\
0 & 1 & 0 & 0 & 0\\0 & 0 &1 & 8 & 0\end{matrix}\right)$$

I let $t$ denote a free variable: $x2=0, x3=-8, x1=2, x4=t$.

I know that the pivot columns must stay in the matrix. I also know that if the amount of vectors is greater than or equal to the span we are still unsure if the vectors are in the given span.

I want to know if I am on the right track and what step I should take next. Can I take out the 4th column and row reduce again to verify that the leftover vectors are in the span?

Best Answer

You don't need an augmented matrix in this case.

Starting with $$A=\left(\begin{matrix}-4 & -7 & 1 & 2\\ 0 & 0 & 3 & 8\\5 & -1 &1 & -4\end{matrix}\right)$$ just row reduce like you did. I'll assume you're correct and that the RREF of this matrix is $$\operatorname{RREF}(A)=\left(\begin{matrix}1 & 0 & 0 & -2\\ 0 & 1 & 0 & 0\\0 & 0 &1 & 8\end{matrix}\right)$$

Then you just count the pivots:

$$\require{enclose}\left(\begin{matrix}\enclose{circle}{1} & 0 & 0 & -2\\ 0 & \enclose{circle}{1} & 0 & 0\\0 & 0 &\enclose{circle}{1} & 8\end{matrix}\right)$$

There are $3$ pivots in this case, meaning the row rank is $3$. By the theorem which tells us the row rank = the column rank of a matrix, we also know that the column rank of $A$ is $3$. Thus there are $3$ linearly independent columns of $A$.

$\Bbb R^3$ has a dimension of $3$ (can you prove this?), thus any $3$ linearly independent vectors will span it. Thus the columns of $A$ do indeed span $\Bbb R^3$.

Appendix: Theorems needed for the above

  1. Row reduction (by elementary row operations) preserves the row space of a matrix
  2. The number of pivots of a matrix in RREF equals the row rank of a matrix.
  3. The row rank = the column rank of a matrix (and thus you can just refer to it as the rank of the matrix).
  4. The dimension of a vector space $V$ is well-defined. I.e. if the dimension of $V$ is $n$, then $(1)$ any $n$ linearly independent vectors $v_1, v_2, \dots, v_n \in V$ is a spanning set, $(2)$ any set of less than $n$ vectors will NOT be a spanning set, and $(3)$ any set of more than $n$ vectors will necessarily be linearly dependent. (So this one is really $3$ different theorems.)

If you haven't encountered one or more of these theorems, you may have to do the problem a different way (or prove these theorems yourself). ;)

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